InterviewSolution
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InterviewSolution
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A force `F = -K(y hatI + x hatj) (where `K` is a posive constant ) acts on a particle moving in the `xy` plane . Starting form the original , the partical is taken along in the positive `x` axis to the point `(x,0)` and then partical to they axis the point `(x,0)` . The total work done by the force `F` on the particls isA. `- 2 Ka^(2)`B. ` 2 Ka^(2)`C. `- Ka^(2)`D. `Ka^(2)` |
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Answer» Correct Answer - C While moving from `(0.0)` to `(a.0)` Along positive x-axis , `y = 0 :. vecF = - kx hatj` i.e. , force is in negative y-direction while dicplacement is in positive direction. `:. W_(1) = 0` Because force is perpendicular to displacement . Then partical moves from `(a.0)` to `(a.a)` during this `vec F = - k(y hati + a hatj)` The first componenet of force, `- ky hatj` will not contribitute any work because this component is along negative x- direction `(- hati)` while displacement is in positive y- direction `(a.0)` to `(a.a)` . The second component of force i.e. `- ka hatj` will perform negative work `:. W_(2) = (- ka hatj) (a hatj) = (-ka) (a) = - ka^(2)` So net work done on the partical `W = W_(1) + W_(2)` `= 0 + (-ka^(2)) = - ka^(2)` |
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